Question

Find the locus of the point P(h, k) if three normals are drawn from the point P to \[{{y}^{2}}=4ax\], satisfying the following \[{{m}_{1}}.{{m}_{2}}=1\].

Answer 1

Hint: First find the equation of the normal which would be a three-degree equation in terms of ‘m’ and put the product of roots = 1. By doing this we will get the required solution.

Complete step-by-step answer:
We are given a parabola \[{{y}^{2}}=4ax\] and a point P(h, k) from which three normals are drawn. Also, we are given that \[{{m}_{1}}{{m}_{2}}=1\] that is the product of slopes of two out of three normals is 1. Now, we have to find the locus of the point P (h, k).
We know that any general point on the parabola \[{{y}^{2}}=4ax\] is \[\left( x,y \right)=\left( a{{t}^{2}},2at \right)\].
We know that any line passing from \[\left( {{x}_{1}},{{y}_{1}} \right)\] and the slope m is:
\[\left( y-{{y}_{1}} \right)=m\left( x-{{x}_{1}} \right)\].
So, the equation of normal at point \[\left( a{{t}^{2}},2at \right)\] and slope m is:
\[\left( y-2at \right)=m\left( x-a{{t}^{2}} \right)....\left( i \right)\]
Now we take the parabola, \[{{y}^{2}}=4ax\].
Now we differentiate the parabola.
\[\left[ \text{Also, }\dfrac{d\left( {{x}^{n}} \right)}{dx}=n{{x}^{n-1}} \right]\]
Therefore, we get,
\[2y\dfrac{dy}{dx}=4a\]
\[\dfrac{dy}{dx}=\dfrac{2a}{y}\]
At \[\left( x,y \right)=\left[ a{{t}^{2}},2at \right]\]
We get, \[\Rightarrow \dfrac{dy}{dx}=\dfrac{2a}{2at}=\dfrac{1}{t}\]
As \[\dfrac{dy}{dx}\] signifies the slope of the tangent, therefore any tangent on parabola at the point \[\left( a{{t}^{2}},2at \right)\] would have a slope \[=\dfrac{1}{t}\].
Now, we know that tangent and normal are perpendicular to each other.
Therefore, \[\left( \text{Slope of tangent} \right)\times \left( \text{Slope of normal} \right)=-1\]
As we have found that \[\text{Slope of tangent =}\dfrac{1}{t}\] and assume that the slope of the normal is m.
Therefore, we get \[\left( \dfrac{1}{t} \right)\times \left( m \right)=-1\].
Hence, t = – m.
Putting the value of t in equation (i), we get,
\[\left[ y-2a\left( -m \right) \right]=m\left[ x-a{{\left( -m \right)}^{2}} \right]\]
\[\Rightarrow y+2am=m\left( x-a{{m}^{2}} \right)\]
\[\Rightarrow y=mx-a{{m}^{3}}-2am\]
By rearranging the given equation, we get,
\[a{{m}^{3}}+m\left( 2a-x \right)+y=0\]
Here, we get a three-degree equation in terms of m, therefore it will have 3 roots \[{{m}_{1}},{{m}_{2}}\] and \[{{m}_{3}}\].
As we know that this normal passes through (h, k), we put x = h and y = k, we get,
\[a{{m}^{3}}+m\left( 2a-h \right)+k=0....\left( ii \right)\]
Comparing the above equation by general three-degree equation \[a{{x}^{3}}+b{{x}^{2}}+cx+d=0\]
We get, \[a=a,\text{ }b=0,\text{ }c=\left( 2a-h \right),\text{ }d=k\]
We know that \[\text{product of roots }=\dfrac{-d}{a}\]
Therefore, we get the product of roots \[=\dfrac{-k}{a}\]
As \[{{m}_{1}},{{m}_{2}}\] and \[{{m}_{3}}\] are the roots,
Hence, \[{{m}_{1}}{{m}_{2}}{{m}_{3}}=\dfrac{-k}{a}\]
As we have been given that \[{{m}_{1}}{{m}_{2}}=1\],
We get \[\left( 1 \right).{{m}_{3}}=\dfrac{-k}{a}\]
Therefore, \[{{m}_{3}}=\dfrac{-k}{a}\]
As \[{{m}_{3}}\]is the root of the equation (ii), therefore, it will satisfy the equation (ii).
Now, we put \[{{m}_{3}}\]in equation (ii), we get
\[a{{\left( {{m}_{3}} \right)}^{3}}+{{m}_{3}}\left( 2a-h \right)+k=0\]
As \[{{m}_{3}}=\dfrac{-k}{a}\]
We get, \[a{{\left( \dfrac{-k}{a} \right)}^{3}}+\left( \dfrac{-k}{a} \right)\left( 2a-h \right)+k=0\]
By simplifying the equation, we get
\[\Rightarrow \dfrac{-{{k}^{3}}}{{{a}^{2}}}-\dfrac{k}{a}\left( 2a-h \right)+k=0\]
\[\Rightarrow \dfrac{-{{k}^{3}}-ka\left( 2a-h \right)+k{{a}^{2}}}{{{a}^{2}}}=0\]
\[\Rightarrow -{{k}^{3}}-ka\left( 2a-h \right)+k{{a}^{2}}=0\]
By taking k common from each term, we get,
\[\left( k \right)\left[ -{{k}^{2}}-a\left( 2a-h \right)+{{a}^{2}} \right]=0\]
\[\Rightarrow \left( k \right)\left[ -{{k}^{2}}-2{{a}^{2}}+ah+{{a}^{2}} \right]=0\]
Now, to get the locus of P(h, k), we will replace h by x and k by y.
Hence, we get
\[\Rightarrow \left( y \right)\left( -{{y}^{2}}+ax-{{a}^{2}} \right)=0\]
\[\Rightarrow -{{y}^{2}}+ax-{{a}^{2}}=0\]
So, the locus of point P(h, k) is \[\left( ax-{{y}^{2}}-{{a}^{2}} \right)=0\].

Note: Kindly take care of negative signs with \[\dfrac{d}{a}\]. Also, many times students confuse with writing the product of roots which is \[-\dfrac{d}{a}\] where d is the constant term of the equation. Also, take care while taking \[{{a}^{2}}\] as the LCM.